We know that there are Whitehead theorem for homotopy and homology theory.
Is there the Whitehead theorem for cohomology theory for 1connected CW complexes?
We know that there are Whitehead theorem for homotopy and homology theory. Is there the Whitehead theorem for cohomology theory for 1connected CW complexes? 


Conceptually, the following two theorems (both due to Whitehead) are EckmannHilton duals. Theorem. A weak homotopy equivalence between CW complexes is a homotopy equivalence. Theorem. A homology isomorphism between simple spaces is a weak homotopy equivalence. They don't look dual, but they are. See
J.P. May.
The dual Whitehead Theorems. The point is that the second statement is really about cohomology, and the standard cellular proof of the first statement dualizes wordforword to a ``cocellular'' proof of the second. Cocellular constructions are what appear in Postnikov towers, and they can be used more systematically than can be found in the literature. Yet another plug: they are central in the upcoming book "More Concise Algebraic Topology'' by Kate Ponto and myself. 


As Sean says, the key point is the Universal Coefficient Theorem, but the details are not completely obvious unless you make some finiteness assumptions. Suppose that $f:X\to Y$ is such that $H^{\ast}(f;\mathbb{Z})$ is an isomorphism. Let $Z$ be the cofibre of $f$, so $\tilde{H}^{\ast}(Z;\mathbb{Z})=0$. If we can prove that $\tilde{H}_{\ast}(Z;\mathbb{Z})=0$ then we can appeal to the ordinary homological Whitehead theorem. MathJax is mangling my tildes: all (co)homology groups of $Z$ below should be read as reduced. For any prime $p$, we have a universal coefficient sequence for $H^{\ast}(Z;\mathbb{Z}/p)$ in terms of $H^{\ast}(Z;\mathbb{Z})$, so $H^{\ast}(Z;\mathbb{Z}/p)=0$. As ${\mathbb{Z}/p}$ is a field we also know that $H_{\ast}(Z;\mathbb{Z}/p)$ is a free module with $H^{\ast}(Z;\mathbb{Z}/p)$ as its dual, so we must have $H_{\ast}(Z;\mathbb{Z}/p)=0$. Using the universal coefficient theorem for homology we deduce that $H_{\ast}(Z;\mathbb{Z})/p$ and $\text{ann}(p,H_{\ast}(Z;\mathbb{Z}))$ are zero, so multiplication by $p$ is an isomorphism on $H_{\ast}(Z;\mathbb{Z})$. As this holds for all $p$, we see that $H_{\ast}(Z;\mathbb{Z})$ is a rational vector space. Thus, if it is nontrivial it will contain a copy of $\mathbb{Q}$ so (via universal coefficients again) $H^{\ast}(Z;\mathbb{Z})$ will contain a copy of $\text{Ext}(\mathbb{Q},\mathbb{Z})$. This group is nonzero (in fact, enormous) by a standard calculation, so this contradicts the initial assumption. 


Sure. The basic point is that for simplyconnected spaces, you can determine the connectivity of a map by looking at the connectivity of the cofiber instead of the connectivity of the fiber. In homology, you determine the connectivity of the cofiber by looking at $H_*(C;\mathbb{Z})$, because of the Hurewicz Theorem. In cohomology, you appeal to: if $X$ is simplyconnected, then $X$ is $n$connected if and only if $[ X, K(G,m)] = *$ for all $m \leq n$ and all abelian groups $G$. This is because (by basic obstruction theory) if $X$ is $(n1)$connected, then $[X, K(G,n)] \cong \mathrm{Hom}(\pi_n(X), G )$. This results in a long list of groups to check, admittedly; but it can be whittled down by Universal Coefficients theorems if you like. 

